Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $a \neq 0$. $y = \dfrac{8a - 40}{a^2 - a} \div \dfrac{a - 5}{a^2 + 3a - 4} $
Dividing by an expression is the same as multiplying by its inverse. $y = \dfrac{8a - 40}{a^2 - a} \times \dfrac{a^2 + 3a - 4}{a - 5} $ First factor the quadratic. $y = \dfrac{8a - 40}{a^2 - a} \times \dfrac{(a - 1)(a + 4)}{a - 5} $ Then factor out any other terms. $y = \dfrac{8(a - 5)}{a(a - 1)} \times \dfrac{(a - 1)(a + 4)}{a - 5} $ Then multiply the two numerators and multiply the two denominators. $y = \dfrac{ 8(a - 5) \times (a - 1)(a + 4) } { a(a - 1) \times (a - 5) } $ $y = \dfrac{ 8(a - 5)(a - 1)(a + 4)}{ a(a - 1)(a - 5)} $ Notice that $(a - 5)$ and $(a - 1)$ appear in both the numerator and denominator so we can cancel them. $y = \dfrac{ 8\cancel{(a - 5)}(a - 1)(a + 4)}{ a\cancel{(a - 1)}(a - 5)} $ We are dividing by $a - 1$ , so $a - 1 \neq 0$ Therefore, $a \neq 1$ $y = \dfrac{ 8\cancel{(a - 5)}\cancel{(a - 1)}(a + 4)}{ a\cancel{(a - 1)}\cancel{(a - 5)}} $ We are dividing by $a - 5$ , so $a - 5 \neq 0$ Therefore, $a \neq 5$ $y = \dfrac{8(a + 4)}{a} ; \space a \neq 1 ; \space a \neq 5 $